Elasticity/Torsion of noncircular cylinders

Torsion of Non-Circular Cylinders edit

Torsion of a noncircular cylinder

Solution first found by St. Venant.
Tractions at the ends are statically equivalent to equal and opposite torques $\pm \mathbf {T} =\pm T{\widehat {\mathbf {e} }}{3}\,$ .
Lateral surfaces are traction-free.

An axis passes through the center of twist ( $x_{3}\,$ axis).
Each c.s. projection on to the $x_{1}-x_{2}\,$ plane rotates,but remains undistorted.
The rotation of each c.s. ( $\phi \,$ ) is proportional to $x_{3}\,$ .

\phi =\alpha x_{3}\,

where $\alpha \,$ is the twist per unit length.

The out-of-plane distortion (warping) is the same for each c.s. and is proportional to $\alpha \,$ .

{\begin{aligned}u_{1}&=r\cos(\phi +\theta )-r\cos \theta =x_{1}(\cos \phi -1)-x_{2}\sin \phi \\u_{2}&=r\sin(\phi +\theta )-r\sin \theta =x_{1}\sin \phi +x_{2}(\cos \phi -1)\\u_{3}&=\alpha \psi (x_{1},x_{2})\end{aligned}}

where $\psi (x_{1},x_{2})\,$ is the warping function.

If $\phi =\alpha x_{3}<<1\,$ (small strain),

{\text{(10)}}\qquad {u_{1}\approx -\alpha x_{2}x_{3}~;~~u_{2}\approx \alpha x_{1}x_{3}~;~~u_{3}=\alpha \psi (x_{1},x_{2})}

\varepsilon _{ij}={\frac {1}{2}}\left(u_{i,j}+u_{j,i}\right)

Therefore,

{\begin{aligned}\varepsilon _{11}&={\frac {1}{2}}\left(0+0\right)=0\\\varepsilon _{22}&={\frac {1}{2}}\left(0+0\right)=0\\\varepsilon _{33}&={\frac {1}{2}}\left(0+0\right)=0\\\varepsilon _{kk}&=\varepsilon _{11}+\varepsilon _{22}+\varepsilon _{33}=0\\\varepsilon _{12}&={\frac {1}{2}}\left(-\alpha x_{3}+\alpha x_{3}\right)=0\\\varepsilon _{23}&={\frac {1}{2}}\left(\alpha \psi _{,2}+\alpha x_{1}\right){\text{(11)}}\qquad \\\varepsilon _{31}&={\frac {1}{2}}\left(\alpha \psi _{,1}-\alpha x_{2}\right){\text{(12)}}\qquad \end{aligned}}

\sigma _{ij}=2\mu \varepsilon _{ij}+\lambda \varepsilon _{kk}\delta _{ij}

Therefore,

{\begin{aligned}\sigma _{11}&=0\\\sigma _{22}&=0\\\sigma _{33}&=0\\\sigma _{kk}&=0\\\sigma _{12}&=0\\\sigma _{23}&=\mu \alpha (\psi _{,2}+x_{1}){\text{(13)}}\qquad \\\sigma _{31}&=\mu \alpha (\psi _{,1}-x_{1}){\text{(14)}}\qquad \end{aligned}}

\sigma _{ji,j}=0~~~~{\text{no body forces.}}

Therefore,

{\begin{aligned}\sigma _{11,1}+\sigma _{21,2}+\sigma _{31,3}=0&\Rightarrow ~~0=0\\\sigma _{12,1}+\sigma _{22,2}+\sigma _{32,3}=0&\Rightarrow ~~0=0\\\sigma _{13,1}+\sigma _{23,2}+\sigma _{33,3}=0&\Rightarrow ~~\mu \alpha (\psi _{,11}+\psi _{,22})=\mu \alpha \nabla ^{2}{\psi }=0{\text{(15)}}\qquad \end{aligned}}

Normal to cross sections is ${\widehat {\mathbf {n} }}{}={\widehat {\mathbf {e} }}{3}\,$ .
Normal traction $t^{n}=\mathbf {t} \bullet {\widehat {\mathbf {n} }}{}=0\,$ .
Projected shear traction is $t^{s}={\sqrt {\sigma _{13}^{2}+\sigma _{23}^{2}}}\,$ .
Traction vector at a point in the cross section is tangent to the cross section.

Lateral surface traction-free.
Unit normal to lateral surface appears as an in-plane unit normal to the boundary $\partial S$ .

We parameterize the boundary curve $\partial S$ using

\mathbf {x} ={\tilde {\mathbf {x} }}(s)~,~~0\leq s\leq l~~;~~~{\tilde {\mathbf {x} }}(0)={\tilde {\mathbf {x} }}(l)

The tangent vector to $s$ is

{\widehat {\boldsymbol {\nu }}}={\frac {d\mathbf {x} }{ds}}~{\text{and}}~~{\widehat {\mathbf {n} }}{}={\widehat {\boldsymbol {\nu }}}\times {\widehat {\mathbf {e} }}_{3}~~\Rightarrow ~~~{\widehat {\mathbf {n} }}{}={\frac {dx_{2}}{ds}}{\widehat {\mathbf {e} }}{1}-{\frac {dx_{1}}{ds}}{\widehat {\mathbf {e} }}_{2}

The tractions $t_{1}\,$ and $t_{2}\,$ on the lateral surface are identically zero. However, to satisfy the BC $t_{3}=0\,$ , we need

t_{3}=n_{1}\sigma _{13}+n_{2}\sigma _{23}=0~~\Rightarrow ~~~\left(\psi _{,1}-x_{2}\right)n_{1}+\left(\psi _{,2}+x_{1}\right)n_{2}=0

or,

{\text{(16)}}\qquad \left(\psi _{,1}-x_{2}\right){\frac {dx_{2}}{ds}}+\left(\psi _{,2}+x_{1}\right){\frac {dx_{1}}{ds}}=0

The traction distribution is statically equivalent to the torque $\mathbf {T} \,$ . At $x_{3}=L\,$ ,

t_{1}=\sigma _{13}~;~~t_{2}=\sigma _{23}~;~~t_{3}=\sigma _{33}=0\,

Therefore,

F_{1}=\int _{S}\sigma _{13}~dS=\mu \alpha \int _{S}(\psi _{,1}-x_{2})~dS

From equilibrium,

{\begin{aligned}\nabla ^{2}{\psi }=0~~\Rightarrow ~~~\psi _{,1}-x_{2}&=(\psi _{,1}-x_{2})+x_{1}(\psi _{,11}+\psi _{,22})\\&=\psi _{,1}+x_{1}\psi _{,11}-x_{2}+x_{1}\psi _{,22}\\&=(x_{1}\psi _{,1}-x_{1}x_{2})_{,1}+(x_{1}\psi _{,2}+x_{1}x_{1})_{,2}\\&=\left[x_{1}(\psi _{,1}-x_{2})\right]_{,1}+\left[x_{1}(\psi _{,2}+x_{1})\right]_{,2}\end{aligned}}

Hence,

{\text{(17)}}\qquad F_{1}=\mu \alpha \int _{S}\left[x_{1}(\psi _{,1}-x_{2})\right]_{,1}+\left[x_{1}(\psi _{,2}+x_{1})\right]_{,2}dS

If $P=f(x_{1},x_{2})\,$ and $Q=q(x_{1},x_{2})\,$ then

\int _{S}(Q_{,1}-P_{,2})dS=\oint _{\partial S}(Pdx_{1}+Qdx_{2})

with the integration direction such that $S$ is to the left.

Applying the Green-Riemann theorem to equation (17), and using equation (16)

{\text{(18)}}\qquad F_{1}=\mu \alpha \oint _{\partial S}-x_{1}(\psi _{,2}+x_{1})dx_{1}+x_{1}(\psi _{,1}-x_{2})dx_{2}=0

Similarly, we can show that $F_{2}=0\,$ . $F_{3}=0\,$ since $t_{3}=0\,$ .

The moments about the $x_{1}\,$ and $x_{2}\,$ axes are also zero.

The moment about the $x_{3}\,$ axis is

M_{3}=\int _{S}(x_{1}\sigma _{23}-x_{2}\sigma _{13})dS=\mu \alpha \int _{S}(x_{1}\psi _{,2}+x_{1}^{2}-x_{2}\psi _{1}+x_{2}^{2})dS=\mu \alpha {\tilde {J}}

where $J\,$ is the torsion constant. Since $M_{3}=T\,$ , we have

\alpha ={\frac {T}{\mu {\tilde {J}}}}

If $\psi =0\,$ , then ${\tilde {J}}=J\,$ , the polar moment of inertia.

Find a warping function $\psi (x_{1},x_{2})\,$ that is harmonic. and satisfies the traction BCs.
Compatibility is not an issue since we start with displacements.
The problem is independent of applied torque and the material properties of the cylinder.
So it is just a geometrical problem. Once $\psi$ is known, we can calculate
- The displacement field.
- The stress field.
- The twist per unit length.